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Automotive Thermodynamics. A Volkswagen Passat has a

University Physics | 13th Edition | ISBN: 9780321675460 | Authors: Hugh D. Young, Roger A. Freedman ISBN: 9780321675460 31

Solution for problem 55P Chapter 20

University Physics | 13th Edition

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University Physics | 13th Edition | ISBN: 9780321675460 | Authors: Hugh D. Young, Roger A. Freedman

University Physics | 13th Edition

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Problem 55P

Automotive Thermodynamics. A Volkswagen Passat has a six-cylinder Otto-cycle engine with compression ratio r = 10.6. The diameter of each cylinder, called the bore of the engine, is 82.5 mm. The distance that the piston moves during the compression in Fig. 20.5, called the ?stroke of the engine, is 86.4 mm. The initial pressure of the air–fuel mixture (at point a in Fig. 20.6) is 8.50 X 104 Pa, and the initial temperature is 300 K (the same as the outside air). Assume that 200 J of heat is added to each cylinder in each cycle by the burning gasoline, and that the gas has C V = 20.5 J / mol ? K and ? = 1.40. (a) Calculate the total work done in one cycle in each cylinder of the engine, and the heat released when the gas is cooled to the temperature of the outside air. (b) Calculate the volume of the air–fuel mixture at point a in the cycle. (c) Calculate the pressure, volume, and temperature of the gas at points b, c, and d in the cycle. In a pV-diagram, show the numerical values of p, V, and T for each of the four states. (d) Compare the efficiency of this engine with the efficiency of a Carnot-cycle engine operating between the same maximum and minimum temperatures.

Step-by-Step Solution:

Solution 55P Step 1 of 4: a)e = 1 1 = 1 1 = 0.6111 r1 10.6.4 e = Q HQ C QH Q C (e 1)Q H = (0.611 1) × 200 J = 78 J W = Q +CQ = H8 J + 200 J = 122 J Step 2 of 4: b) The stoke times the bore equals the change in volume. The initial volume is the final volume V times the compression ratio r.Combining these two expressions gives an equation for V . For each cylinder of area 2 d 2 A = r = ( ) 2 the piston moves 0.864 m and the volume changes from rV to V , as shown in figure. 1 A = rV l A = V 2 3 (1 l2)=86.4× 10 m l A l A = rV V 1 2 A(l 1 l 2 = V (r 1) A(l l ) V = 1 2 r1 2 = r (1 2 ) r1 3 3 2 = 3.14×86.4×10 m×(1.25×10 ) 10.61 5 3 = 4.811 × 10 m

Step 3 of 4

Chapter 20, Problem 55P is Solved
Step 4 of 4

Textbook: University Physics
Edition: 13
Author: Hugh D. Young, Roger A. Freedman
ISBN: 9780321675460

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